Mastering physics - speed of a softball, help please!?

Question

A softball is hit over a third baseman's head with some speed v_0 at an angle theta above the horizontal. Immediately after the ball is hit, the third baseman turns around and begins to run at a constant velocity V = 7.00m/s. He catches the ball t=2.00s later at the same height at which it left the bat. The third baseman was originally standing L=18.0m from the location at which the ball was hit. Find v_0. Use g=9.81 m/s^2 for the magnitude of the acceleration due to gravity.

Answer 1

It would be nice if you could give us the formula next time.

Answer 2

Range = VocosΘ*t = 18+2*7 = 32 m

Since the ball goes up for 1 sec, then down for 1sec,

Vvi = Vo*sinΘ = g*1sec

Solving these 2 for Θ & Vo → Θ = 31.48° and Vo = 18.77 m/s